In a C*-algebra, pure states which share the same kernel are equal

Question

I'm reading C*-Algebras by Jacques Dixmier. And in the proof of 2.9.5, it says

Let $A$ be a C*-algebra. If $f$ and $f'$ are two pure states which have the same kernel, then $f=f'$.

It should be obvious I guess, but I cannot understand why. It would be very helpful if anyone can give me some hint.

Answer 1

It has nothing to do with pure, not even with positivity. States extend uniquely to the unitization of $A$, so we may assume that $A$ is unital .

Let $x\in A$. Then $x-f(x)I\in \ker f=\ker f'$. Then $$ 0=f'(x-f(x)I)=f'(x)-f(x). $$ So all we are using is linearity and that $f(I)=f'(I)=1$. For two arbitrary functional we the same kernel, we would obtain $f'=c\,f$ for some constant $c$.

As for books, the canonical sources are Murphy's book, and Davidson's C$^*$-algebras by Example.